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Math  /  Data & Statistics

QuestionA true/false test has 90 questions. Suppose a passing grade is 54 or more correct answers. Test the claim that a student knows more than half of the answers and is not just guessing. Assume the student gets 54 answers correct out of 90 . Use a significance level of 0.05 . Steps 1 and 2 of a hypothesis test procedure are given below. Show step 3 , finding the test statistic and the p-value and step 4 , interpreting the results.
Step 1: H0:p=0.50H_{0}: p=0.50 Ha:p>0.50H_{a}: p>0.50
Step 2: Choose the one-proportion z-test. Sample size is large enough, because np0n p_{0} is 90(0.5)=4590(0.5)=45 and n(1−p0)=n\left(1-p_{0}\right)= 90(.5)=4590(.5)=45, and both are more than 10 . Assume the sample is random. Step 3: Compute the zz-test statistic, and the pp-value. z=z= □\square (Round to two decimal places as needed.) pp-value == □\square (Round to three decimal places as needed.)

Studdy Solution

STEP 1

What is this asking? Did our student *actually* study, or did they just get lucky guessing?
We're checking if their score of 54/90 is *significantly* better than just a 50% guess. Watch out! Don't mix up the *null* hypothesis (just guessing) with what we're actually testing (did they study?).

STEP 2

1. Calculate the Z-score
2. Calculate the p-value
3. Interpret the results

STEP 3

The z-score formula is like a magical detector for how unusual an event is.
It tells us how far our student's score is from the average guessing score, measured in "standard deviations".
Here's the formula: z=p^−p0p0(1−p0)n z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} Where: p^\hat{p} is the **sample proportion** (what our student *actually* scored), p0p_0 is the **hypothesized proportion** (assuming they were guessing), and nn is the **sample size** (number of questions).

STEP 4

Let's plug in what we know! p^=5490=0.6\hat{p} = \frac{54}{90} = 0.6, p0=0.5p_0 = 0.5 (because we're assuming a 50% guess rate), and n=90n = 90. z=0.6−0.50.5(1−0.5)90 z = \frac{0.6 - 0.5}{\sqrt{\frac{0.5(1-0.5)}{90}}}

STEP 5

First, simplify the numerator and denominator separately: 0.6−0.5=0.10.6 - 0.5 = 0.1 and 0.5(1−0.5)=0.5⋅0.5=0.250.5(1-0.5) = 0.5 \cdot 0.5 = 0.25. z=0.10.2590 z = \frac{0.1}{\sqrt{\frac{0.25}{90}}} Next, divide 0.250.25 by 9090 to get approximately 0.0027780.002778. z=0.10.002778 z = \frac{0.1}{\sqrt{0.002778}} Now, take the square root of 0.0027780.002778, which is approximately 0.05270.0527. z=0.10.0527 z = \frac{0.1}{0.0527} Finally, divide 0.10.1 by 0.05270.0527 to get our **z-score**, which is approximately 1.901.90.

STEP 6

The p-value tells us how likely it is to get a result *as extreme* as our student's score (or even *more* extreme) if they were *just guessing*.
A small p-value means it's *unlikely* they were guessing.

STEP 7

We need to find the area to the *right* of our **z-score** (1.901.90) on the standard normal distribution.
Using a z-table or calculator, we find that the area to the *left* of 1.901.90 is approximately 0.97130.9713.
Since the total area under the curve is 11, the area to the *right* (our **p-value**) is 1−0.9713=0.02871 - 0.9713 = 0.0287.
Rounded to three decimal places, our **p-value** is **0.029**.

STEP 8

Our **p-value** (**0.029**) is *less than* our **significance level** (0.050.05).

STEP 9

This means it's *unlikely* our student just got lucky guessing.
We *reject* the null hypothesis (that they were just guessing) and conclude that there's evidence to suggest they *do* know more than half the answers!

STEP 10

z=1.90z = 1.90 p-value=0.029p\text{-value} = 0.029

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